leetcode题解-36.Valid Sudoku

题目：Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character ‘.’. A partially filled sudoku which is valid.

Note:

A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

其实就是检验数独棋盘中已经给出的数字是否满足数独规则。即每行、每列、每个小九宫格内的数字是否重复。

思路：遍历棋盘，分别考察行、列、九宫格内数字是否满足条件，共九行，九列，九个九宫格。

1，使用三个hashset分别存储三个维度

public boolean isValidSudoku1(char[][] board) {
for(int i=0; i<9; i++)
{
Set<Character> row = new HashSet<Character>();
Set<Character> col = new HashSet<Character>();
Set<Character> cube = new HashSet<Character>();
for(int j=0; j<9; j++)
{
if(board[i][j] != '.')
if(!row.add(board[i][j]))
return false;
if(board[j][i] != '.')
if(!col.add(board[j][i]))
return false;
int RowIndex = 3*(i/3);
int ColIndex = 3*(i%3);
if(board[RowIndex + j/3][ColIndex + j%3] != '.')
if(!cube.add(board[RowIndex+j/3][ColIndex + j%3]))
return false;
}
}
return true;
}

2，使用数组存储。这种方法速度更快。

public boolean isValidSudoku(char[][] board) {
int [] vset = new int [9];
int [] hset = new int [9];
int [] bckt = new int [9];
int idx = 0;
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if (board[i][j] != '.') {
idx = 1 << (board[i][j] - '0') ;
if ((hset[i] & idx) > 0 ||
(vset[j] & idx) > 0 ||
(bckt[(i / 3) * 3 + j / 3] & idx) > 0) return false;
hset[i] |= idx;
vset[j] |= idx;
bckt[(i / 3) * 3 + j / 3] |= idx;
}
}
}
return true;
}